# Fractional Exponential

1. Nov 12, 2009

### Sirsh

I have been given a Indice. I've been trying to figure it out for awhile and need some assistance, It'd be great if someone could work it out and show me the steps they did and even explain it.

([a-2b-3/ 2a3b-4]2 )/ [ab-15]/a-3b2])

I've attempted the working out in my book, Its abit hard to put the whole working on here as it's time consuming. I'm down to [4a^2b^2]/[a^-12b]

thank you.

2. Nov 12, 2009

### Staff: Mentor

$$\left[{\frac{a^{-2}b^{-3}}{2a^3b^{-4}}\right]^2~\frac{a^{-3}b^2}{ab^{-15}}$$

One approach would be to move all the factors with negative exponents from the top to the bottom or from the bottom to the top in each fraction, which would leave you with positive exponents. The idea is that a-n = 1/an. After that, square each factor in the first fraction. Finally, cancel everything that can be cancelled.

3. Nov 12, 2009

### Sirsh

Thanks for that :) is this how i do it?

[a-2b-3 / 2a3b-4 ]2 / [a-3b2 / ab-15]

[2a3b4 / a2b3 ]2 / [ab15 / a3b2 ]

[4a6b8 / a4b6 ] = [4a2b2 ]

and [ab15 / a3b2 ] = [a12b13 ]

I am unsure about the second half.

4. Nov 12, 2009

### icystrike

$$(\frac{a^{-2}b^{-3}}{2a^{3}b^{-4}})(\frac{a^{-3}b^{2}}{ab^{-15}})$$
First take a look at the left side;

It will give
$$\frac{a^{(-2-3)}b^{[-3-(-4)]}}{2}$$
Which is equal to
$$\frac{a^{-5}b}{2}$$

Now look at right side;
$$a^{(-3-1)}b^{[2-(-15)]}$$
$$a^{-4}b^{13}$$

Combine both LHS and RHS;
$$\frac{a^{[-5+(-4)]}b^{(1+1)}}{2}$$

I will leave the rest to you, noting that $$a^{-9} = \frac{1}{a^{9}}$$

5. Nov 12, 2009

### Sirsh

Did you neglect to use the square of the left side? I dont see where you used it.

6. Nov 12, 2009

### icystrike

opps.. missed it.
then i guess you just have to square LHS before combining.

7. Nov 12, 2009

### Sirsh

That's okay :) and btw, the LFH is divided by the RHS. sorry for confusion I dont know how to work the script stuff on this site.

8. Nov 12, 2009

### Sirsh

[ $$\frac{a^-^2b^-^3}{2a^3b^-^4}$$ ]2 / $$\frac{ab^-^1^5}{a^-3b^2}$$

$$\frac{a^-^4b^-^6}{4a^6b^-8}$$ / $$\frac{ab^-^1^5}{a^-^3b^2}$$

$$\frac{a^(^-^4^-^6^)b^(^-^6^-^(^-^8^)^)}{4}$$ / $$\frac{a^(^-^1^-^3^)b^(^-^1^5^-^2^)}$$

$$\frac{a^-^1^0b^2}{4}$$ / $$\frac{a^2b^-^1^7}$$

$$\frac{a^(^-^1^0^-^2^)b^(^2^-^(^-^1^7^)}{4}$$

$$\frac{a^-^1^2b^1^9}{4}$$

I ended up getting $$\frac{4}{a^1^2b^1^9}$$

If you ended up trying it, did you also get that?

9. Nov 12, 2009

### icystrike

I see mistakes in the second expression

10. Nov 12, 2009

### Sirsh

Could you point them out please, I'm not very good at indices.

11. Nov 12, 2009

### Sirsh

$$\frac{b^1^9}{4a^1^2}$$

The answer is $$\frac{b^1^9}{4a^1^4}$$

So, now the only thing I need to find out is where i lost a 2.

12. Nov 12, 2009

### caocuong93

Hi. I'm just a newbie but would like to give it a try.

Look for any mistakes !!!

This is what I got after a while working it out:

$$\left[{\frac{a^{-2}b^{-3}}{2a^3b^{-4}}\right]^2*~\frac{a^{-3}b^2}{ab^{-15}} = \frac{\left(a^{-2}b^{-3}\right)^2}{\left(2a^3b^{-4}\right)^2}\right]*~\frac{a^{-3}b^2}{ab^{-15}} = {\frac{a^{-4}b^{-6}}{4a^6b^{-8}}~*~\frac{a^{-3}b^2}{ab^{-15}} = {\frac{a^{-4-3}b^{-6+2}}{4a^{6+1}b^{-8-15}} = {\frac{a^{-7}b^{-4}}{4a^{7}b^{-23}} = {\frac{b^{-4+23}}{4a^{7+7}} = {\frac{b^{19}}{4a^{14}}$$

By the way, I wonder what class you're taking, 'cause my Pre-Cal class hasn't had this kind of stuff yet.

CN

Last edited: Nov 12, 2009
13. Nov 12, 2009

### caocuong93

Double-posted, sorry! I don't know how to delete doubled posts.

14. Nov 12, 2009

### Anakin_k

Nope, no mistakes. I'm in Pre-Calc at the moment too, none of this stuff so far. We worked on this kind of stuff in grade 11 if memory serves me right.

15. Nov 13, 2009

### Sirsh

Thanks guys for the help :) I'm from australia and i'm currently doing the Introcalculus course in year 11. I use to do the Calculus one but my study load was to much.